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%%文档的题目、作者与日期

%\author{王立庆（2019级数学与应用数学1班）}
\author{学号 \underline{\hspace{4cm}} 姓名  \underline{\hspace{4cm}} }
%\title{高等代数第六章：向量空间}
\title{第七章线性变换（7.4-7.6）考试}
%\date{\vspace{-3ex}}
\renewcommand{\today}{\number\year \,年 \number\month \,月 \number\day \,日}
\date{2023年4月19日}

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\begin{document}

\maketitle

\thispagestyle{empty}

%\begin{abstract}
%%主要内容：
%7.3. 
%7.4. 
%7.5. 

%
%\end{abstract}

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\begin{enumerate}

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\item %1
设 $V=\mathbb{R}[x]_3$ 是次数小于等于 3 的多项式与零多项式组成的向量空间。设线性变换 $\sigma\in L(V)$ 由如下定义，
$$\sigma(f(x)) = xf'(x)+2f(x).$$
\begin{enumerate}
\item 验证 $U=L(x^2)$ 是 $\sigma$ 的一个不变子空间。
\item 验证 $W=L(1,x)$ 也是 $\sigma$ 的一个不变子空间。
%\item 你还能找出 $\sigma$ 的其它不变子空间吗？
\item 写出 $\sigma$ 关于基 $\{1,x,x^2,x^3\}$ 的矩阵。
\end{enumerate}

%\vspace{0.2cm}
%
%{\color{red}解答：
%
%\begin{enumerate}
%\item 对任意 $f(x)=ax^2\in U$, 有 $\sigma(f(x))=x\cdot 2ax+2ax^2=4ax^2\in U$, 所以 $\sigma(U)\subseteq U$.
%\item 对任意 $f(x)=a+bx\in W$, 有 $\sigma(f(x))=x\cdot b+2(a+bx)=2a+3bx\in W$, 所以 $\sigma(W)\subseteq W$. 
%%\item 由 $\{1,x,x^2,x^3\}$ 的任意子集张成的子空间都是这个 $\sigma$ 的不变子空间。
%\item 根据线性变换关于一个基的矩阵的定义，写出下述等式，
%\begin{eqnarray*}
%(\sigma(1), \sigma(x), \sigma(x^2), \sigma(x^3)) = (2, 3x, 4x^2, 5x^3) = 
%(1,x,x^2,x^3)\begin{pmatrix} 2&0&0&0 \\  0&3&0&0 \\  0&0&4&0 \\  0&0&0&5 \\   \end{pmatrix}.
%\end{eqnarray*}
%右边的四阶矩阵即为所求。
%
%\end{enumerate}
%
%}

\vspace{0.2cm}

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\item %2
计算矩阵 $A=\begin{pmatrix} 2&1 \\ 2&3 \end{pmatrix}$ 的特征多项式、特征值和特征向量。

%\vspace{0.2cm}
%
%{\color{red}解答：
%
%\begin{enumerate}
%\item[(1)]  首先计算特征多项式，
%\begin{eqnarray*}
%f(\lambda) = \det(\lambda E-A) = \begin{vmatrix} \lambda -2 &-1 \\ -2&\lambda-3 \end{vmatrix} =  (\lambda -2)(\lambda -3) -2 
%= \lambda^2-5\lambda +4. 
%\end{eqnarray*}
%
%\item[(2)]  然后计算特征值，从 $\lambda^2-5\lambda +4=0$ 可得 $\lambda_1=1$, $\lambda_2=4$. 
%
%\item[(3)]  最后对每个特征值计算相应的特征向量，
%\begin{enumerate}
%\item  从 $(\lambda_1E-A)X=0$ 可得 $X=k(1,-1)^t$, 其中 $k\in\mathbb{R}, k\neq 0$.
%\item  从 $(\lambda_2E-A)X=0$ 可得 $X=k(1,2)^t$, 其中 $k\in\mathbb{R}, k\neq 0$.
%\end{enumerate}
%
%\end{enumerate}
%
%}

\vspace{0.2cm}

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\item %3
设 $V$ 是二维实向量空间，设 $\Phi=\{\alpha_1,\alpha_2\}$ 是 $V$ 的一个基。设 $\sigma\in L(V)$ 是一个线性变换。设
\begin{eqnarray*}
\sigma(\alpha_1) &=& 2\alpha_1 + 2\alpha_2, \\
\sigma(\alpha_2) &=& \alpha_1 + 3\alpha_2. 
\end{eqnarray*}
求 $V$ 的另一个基 $\Psi=\{\beta_1,\beta_2\}$, 使得 $\sigma$ 关于 $\Psi$ 的矩阵是对角阵。

%\vspace{0.2cm}
%
%{\color{red}解答：
%
%\begin{enumerate}
%
%\item[(1)]  
%从题目条件可知 $\sigma$ 关于基 $\Phi$  的矩阵是下述等式右边的二阶矩阵， 
%\begin{eqnarray*}
%(\sigma(\alpha_1), \sigma(\alpha_2)) = (\alpha_1,\alpha_2)\begin{pmatrix} 2&1 \\ 2&3 \end{pmatrix}.
%\end{eqnarray*}
%
%\item[(2)]  
%由上一题可知，矩阵 $A=\begin{pmatrix} 2&1 \\ 2&3 \end{pmatrix}$ 有特征值 $\lambda_1=1$ 与 $\lambda_2=4$, 以及相应的特征向量 $(1,-1)$ 与 $(1,2)$. 
%
%\item[(3)]  
%所以取 
%\begin{eqnarray*}
%\beta_1 &=& \alpha_1-\alpha_2, \\   
%\beta_2 &=& \alpha_1+2\alpha_2, 
%\end{eqnarray*}
%则可以验证 $\{\beta_1,\beta_2\}$ 是一个基，而且有 
%\begin{align*}
%\sigma(\beta_1) &= \sigma(\alpha_1-\alpha_2) = \sigma(\alpha_1)-\sigma(\alpha_2) = (2\alpha_1 + 2\alpha_2) - (\alpha_1 + 3\alpha_2) = \alpha_1-\alpha_2  = \beta_1, \\
%\sigma(\beta_2) &= \sigma(\alpha_1+2\alpha_2) = \sigma(\alpha_1)+2\sigma(\alpha_2)= (2\alpha_1 + 2\alpha_2) + 2(\alpha_1 + 3\alpha_2) = 4(\alpha_1+2\alpha_2) = 4\beta_2.  
%\end{align*}
%
%\item[(4)]  
%从而有 
%\begin{eqnarray*}
%(\sigma(\beta_1), \sigma(\beta_2)) = (\beta_1,\beta_2)\begin{pmatrix} 1&0 \\ 0&4 \end{pmatrix}, 
%\end{eqnarray*}
%即 $\sigma$ 关于基 $\Psi=\{\beta_1,\beta_2\}$ 的矩阵是对角阵。
%
%\end{enumerate}
%
%}

\vspace{0.2cm}

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\end{enumerate}



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